Calculate the charge to mass ratio in this mass spectrometer problem

In summary: Do you know where the formula “q/m= 2v/B^2 R^2“comes from? Is it a...I can't answer that question without more information about the problem and the class you are in.
  • #1
kconnolly
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Homework Statement
In a mass spectrometer, a charged particle is first accelerated in an electric field through a potential difference of 1.80 kV. It exits the electric field and enters a uniform magnetic field, perpendicular to the field direction. The magnetic field strength is 0.038 T when the particle accelerates with a values of 7.28E+08 m/s². Assume all the electrical potential energy of the particle is converted to kinetic energy when the particle is accelerated in the electric field.

(a) Determine the charge to mass ratio of the particle.

C/kg

(b) If the particle is a singly charged positive ion, calculate its mass.
Relevant Equations
i think i could me q/m= 2v/B^2 R^2
2(7.28E+08) / (0.038^2) R^2
i don't know how to get r
 
Last edited:
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  • #3
kconnolly said:
Homework Statement:: In a mass spectrometer, a charged particle is first accelerated in an electric field through a potential difference of 1.80 kV. It exits the electric field and enters a uniform magnetic field, perpendicular to the field direction. The magnetic field strength is 0.038 T when the particle accelerates with a values of 7.28E+08 m/s². Assume all the electrical potential energy of the particle is converted to kinetic energy when the particle is accelerated in the electric field.

(a) Determine the charge to mass ratio of the particle.

C/kg

(b) If the particle is a singly charged positive ion, calculate its mass.
Relevant Equations:: no clue

dont know what to do
As mentioned, you need to show us the Relevant Equation(s) and show your initial work on the problem before we can be of much tutorial help.

That said, you are probably studying the Lorentz Force, correct...?
 
  • #4
kconnolly said:
Homework Statement:: In a mass spectrometer, a charged particle is first accelerated in an electric field through a potential difference of 1.80 kV. It exits the electric field and enters a uniform magnetic field, perpendicular to the field direction. The magnetic field strength is 0.038 T when the particle accelerates with a values of 7.28E+08 m/s². Assume all the electrical potential energy of the particle is converted to kinetic energy when the particle is accelerated in the electric field.

(a) Determine the charge to mass ratio of the particle.

C/kg

(b) If the particle is a singly charged positive ion, calculate its mass.
Relevant Equations:: i think i could me q/m= 2v/B^2 R^2

2(7.28E+08) / (0.038^2) R^2
i don't know how to get r
I see you've added some work into your original post. Have you worked with the vector Lorentz Force yet? It's easiest to work this problem using vectors, IMO.
 
  • #5
berkeman said:
I see you've added some work into your original post. Have you worked with the vector Lorentz Force yet? It's easiest to work this problem using vectors, IMO.
I don't really know to be honest. I am quiet lost on how to do it. don't even really know how to start it
 
  • #6
What are you studying now that prompted this homework question? Are you using vectors in your calculations in that class, or just trig formulas? What reading have you been doing about the Lorentz Force?
 
  • #7
berkeman said:
What are you studying now that prompted this homework question? Are you using vectors in your calculations in that class, or just trig formulas? What reading have you been doing about the Lorentz Force?
`i am doing magnetism. i think we are just using trig formulas. what i have gotten so far is
A->q/m =2v / B^2 R^2
2(7.28*10^8)/ (0.038^2) R^2
then use r =mv/qb to get r
B-> 1.6*10^-19
i just don't know how to actually get r
 
  • #8
or could you use q/m = .5(velocity^2) / voltage
 
  • #9
I'm not able to follow your equations without starting at the beginning. Here is what the Lorentz Force looks like as a vector equation:

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html
1618758126240.png


It says that the force on a charged particle q due to electric field E is qE and is in the same axis as the direction of E. It also says that the magnetic force on q is qv X B (where "X" is the vector cross product), so that force is perpendicular to both the velocity v and the direction of the magnetic field B.

So if the charged particle is traveling at a right angle to B, then the magnetic force equation reduces to F = qvB, and is perpendicular to both the velocity v and the direction of the B field.

The acceleration that they are giving for the charged particle is due to the centripital acceleration of the particle around the magnetic field lines. You figure out the velocity of the particle based on its initial Kinetic Energy (KE) from the 1.80keV energy it got accelerating in the electric field part of the mass spectrometer.

What is the equation for the acceleration of a particle in uniform circular motion in terms of its mass and velocity and the radius of the circular motion? What is the units conversion from energy in keV to Joules?
 
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  • #10
kconnolly said:
A->q/m =2v / B^2 R^2
2(7.28*10^8)/ (0.038^2) R^2
then use r =mv/qb to get r
B-> 1.6*10^-19
i just don't know how to actually get r
You have substituted "7.28*10^8" for the value for v in your equation. But 7.28*10^8 m/s² is an acceleration, not a speed. You can’t do that!

Do you know where the formula “q/m= 2v/B^2 R^2“comes from? Is it a standard formula from your data sheet? (If not, you shouldn't use it.)

There is no need to find/use the radius of motion. Try this approach:

a) Derive an expression for the particle’s speed when it exits the electric field

b) Write an expression for the magnetic force experienced by the particle in the magnetic field.

c) Apply ‘F=ma’ to come up with an equation from which (with some algebra) q/m can be found.

For information, I think there is a mistake in the original question.
“[The particle] exits the electric field and enters a uniform magnetic field perpendicular to the field direction.”
should say:
“[The particle] exits the electric field and enters a uniform magnetic field which has a direction perpendicular to the particle’s velocity.”

Edited - typo'.
 
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  • #11
sorry i really don't understand the other ways
 
  • #12
berkeman said:
I'm not able to follow your equations without starting at the beginning. Here is what the Lorentz Force looks like as a vector equation:

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html
View attachment 281729

It says that the force on a charged particle q due to electric field E is qE and is in the same axis as the direction of E. It also says that the magnetic force on q is qv X B (where "X" is the vector cross product), so that force is perpendicular to both the velocity v and the direction of the magnetic field B.

So if the charged particle is traveling at a right angle to B, then the magnetic force equation reduces to F = qvB, and is perpendicular to both the velocity v and the direction of the B field.

The acceleration that they are giving for the charged particle is due to the centripital acceleration of the particle around the magnetic field lines. You figure out the velocity of the particle based on its initial Kinetic Energy (KE) from the 1.80keV energy it got accelerating in the electric field part of the mass spectrometer.

What is the equation for the acceleration of a particle in uniform circular motion in terms of its mass and velocity and the radius of the circular motion? What is the units conversion from energy in keV to Joules?
i1.8kev = 2.8839E-16 J
i don't really know how to do the velocity thing
would it be v^2= ke/.5m
 
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  • #13
kconnolly said:
i1.8kev = 2.8839E-16 J
i don't really know how to do the velocity thing
would it be v^2= ke/.5m
Yes, ##KE = \frac{1}{2}m v^2##

(note that I used in-line LaTeX to generate that math equation -- see the "LaTeX Guide" in the lower left of the Edit window)

And what is the equation for the centripital force in uniform circular motion involving mass, velocity and radius?
 

Related to Calculate the charge to mass ratio in this mass spectrometer problem

1. What is the purpose of calculating the charge to mass ratio in a mass spectrometer problem?

The charge to mass ratio is an important measurement in a mass spectrometer because it helps identify the type and structure of molecules present in a sample. It also helps in determining the atomic mass of elements and their isotopes.

2. How is the charge to mass ratio calculated in a mass spectrometer problem?

The charge to mass ratio is calculated by measuring the amount of deflection of charged particles in a magnetic field. This deflection is dependent on the charge and mass of the particles, and by measuring the deflection and knowing the strength of the magnetic field, the charge to mass ratio can be determined.

3. What is the unit of measurement for charge to mass ratio in a mass spectrometer problem?

The unit of measurement for charge to mass ratio is Coulombs per kilogram (C/kg) in the SI system. However, it can also be measured in other units such as electron volts per atomic mass unit (eV/amu) or Farads per kilogram (F/kg).

4. How does the charge to mass ratio affect the results in a mass spectrometer?

The charge to mass ratio plays a crucial role in the accuracy and precision of the results obtained in a mass spectrometer. An incorrect charge to mass ratio can lead to incorrect identification of molecules and their isotopes, resulting in inaccurate data.

5. Are there any limitations to using the charge to mass ratio in a mass spectrometer?

While the charge to mass ratio is a valuable measurement in a mass spectrometer, it has limitations. It assumes that all particles have the same charge and only one type of charge, which may not always be the case. Additionally, it does not account for the presence of neutral particles in the sample.

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